The radius of a solid sphere and a solid right circular cone are equal and the height of the cone is equal to the diameter of its base. They are melted and the combined matter is recast into a solid hemisphere. What is the ratio of the radius of the hemisphere to that of the circle?

Let's denote the radius of the sphere and the base of the cone as r. The height of the cone is then 2r (since it's equal to the diameter of the base).

The volume of the sphere is given by the formula (4/3)πr³, and the volume of the cone is given by the formula (1/3)πr²h = (1/3)πr²(2r) = (2/3)πr³.

When the sphere and the cone are melted and recast into a hemisphere, the volume of the hemisphere is equal to the sum of the volumes of the sphere and the cone.

So, the volume of the hemisphere is (4/3)πr³ + (2/3)πr³ = 2πr³.

The volume of a hemisphere is given by the formula (2/3)πR³, where R is the radius of the hemisphere. Setting this equal to the volume of the combined matter, we get:

(2/3)πR³ = 2πr³.

Solving for R, we get R = r√2.

So, the ratio of the radius of the hemisphere to that of the sphere (or the base of the cone) is √2:1.

Let's denote the radius of the sphere and the base of the cone as r. The height of the cone is then 2r (since it's equal to the diameter of the base).

The volume of the sphere is given by the formula (4/3)πr³.

The volume of the cone is given by the formula (1/3)πr²h = (1/3)πr²(2r) = (2/3)πr³.

When the sphere and the cone are melted together, the total volume is (4/3)πr³ + (2/3)πr³ = 2πr³.

This is then recast into a hemisphere. The volume of a hemisphere is given by the formula (2/3)πR³, where R is the radius of the hemisphere.

Setting this equal to the total volume, we get (2/3)πR³ = 2πr³.

Solving for R, we get R = r√2.

Therefore, the ratio of the radius of the hemisphere (R) to that of the sphere (r) is √2:1.